查找差距序列号 [英] Finding gaps sequential numbers

问题描述

我不为生活所以请原谅我，如果它是一个简单的问题（或更复杂的比我想）做这个东西。我已经挖通过档案，发现有很多是接近技巧但作为一个新手，我不知道如何调整我的需要，或他们是远远超出我的理解。

I don’t do this stuff for a living so forgive me if it’s a simple question (or more complicated than I think). I‘ve been digging through the archives and found a lot of tips that are close but being a novice I’m not sure how to tweak for my needs or they are way beyond my understanding.

我有一些大的数据文件，我可以解析出生成坐标大多顺序

I have some large data files that I can parse out to generate a list of coordinate that are mostly sequential

5
6
7
8
15
16
17
25
26
27

我要的是空白的列表

What I want is a list of the gaps

1-4
9-14
18-24

我不知道的Perl，SQL或任何幻想，但我想也许可以做一些事情，会从下一个减去一个数。然后，我至少可以用grep的输出，其中差异无1或-1，并与合作得到的空白。

I don’t know perl, SQL or anything fancy but thought I might be able to do something that would subtract one number from the next. I could then at least grep the output where the difference was not 1 or -1 and work with that to get the gaps.

推荐答案

通过的 AWK ：

awk '$1!=p+1{print p+1"-"$1-1}{p=$1}' file.txt

说明

• $ 1 是当前输入行的第一列


• P 是最后一行的previous值


• 所以（$ 1 = P + 1！）是一个条件：如果 $ 1 低于$ P $不同pvious值+1，则：


• 这部分被执行： {打印P + 1 - $ 1-1} ：打印previous值+1，在 - 字符握拳柱+ 1


• {P = $ 1} 为每个行执行： P 被分配到当前第1列



explanations

• $1 is the first column from current input line
• p is the previous value of the last line
• so ($1!=p+1) is a condition : if $1 is different than previous value +1, then :
• this part is executed : {print p+1 "-" $1-1} : print previous value +1, the - character and fist columns + 1
• {p=$1} is executed for each lines : p is assigned to the current 1st column

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